A critiria of measurability related to Lebesgue Density Theorem

lebesgue-measurereal-analysis

The statement is as follows.

Assume $E\subset\mathbb{R}$, $\forall x\in E$, $$\lim_{\delta\to0}\frac{m^*(E^c\cap (x-\delta,x+\delta))}{2\delta}=0$$
where $E^c$ denotes the complement of $E$. Can we conclude that $E$ is a Lebesgue measurable set? $m$ denotes the standard Lebesgue measure and $m^*$ denotes the outer Lebesgue measure on $\mathbb R$.

Any suggestion is welcomed. Thanks a lot!

Best Answer

The answer is yes, $E$ is measurable. Here is a theorem that holds for $\mathbb{R}^n$

Theorem: Let $E\subset \mathbb{R}^n$. Then $$\begin{align} \lim_{\delta\rightarrow0}\frac{m^*(E\cap B(x;\delta))}{m(B(x;\delta))}=1\tag{1}\label{one} \end{align}$$ for almost all $x\in E$. Furthermore, $E$ is Lebesgue measurable iff $$\begin{align} \lim_{\delta\rightarrow0}\frac{m^*(E\cap B(x;\delta))}{m(B(x;\delta))}=0\tag{2}\label{two} \end{align}$$ for almost all $x\in E^c$.

A proof of this result can be found in Jones, F. Lebesgue Integral on Euclidean space, Revised edition, Jone & Bartlett, 2001. p. 464

Here is an outline to the proof: To simplify arguments, I consider only the case where $m^*(E)<\infty$. Let $A$ a Borel set such that $E\subset A$ and $m^*(E)=m(A)$. (Take for example open sets $E\subset A_n$ such that $m(A_n)<m^*(E)+2^{-n}$ and define $A=\bigcap_nA_n$).

The following fact reduces the problem to that of a measurable set:

Prop 1. For any Lebesgue measurable set $B$, \begin{align} m^*(E\cap B)=m(A\cap B)\tag{0}\label{zero} \end{align}

A proof is provided at the end of this posting.

From Proposition 1 it follows that $$\frac{m^*(E\cap B(x;\delta))}{m(B(x;\delta))}=\frac{m(A\cap B(x;\delta))}{m(B(x;\delta))}$$ Therefore, by Lebesgue's theorem $$\begin{align}\lim_{\delta\rightarrow0}\frac{m^*(E\cap B(x;\delta))}{m(B(x;\delta))}=\lim_{\delta\rightarrow0}\frac{m(A\cap B(x;\delta))}{m(B(x;\delta))}=1_A(x)\qquad \text{a.a}\,x\in\mathbb{R}^n\tag{3}\label{three} \end{align}$$ Since $E\subset A$, \eqref{one} follows, as well as \eqref{two} when $E$ is Lebesgue measurable.

Conversely, assume that \eqref{two} holds for all $x\in E^c$. By \eqref{three}, $A\cap E^c$ is a null set and so Lebesgue measurable; hence $E=A\setminus (A\cap E^c)$ is Lebesgue measurable.

Edit: In response to a comment from @FShrike, and to provide a proof for \eqref{zero}.

For any set $E\subset \mathbb{R}^n$, define $m_*(A)=\sup\{m(K): K\subset E, \, K\,\text{compact}\}$. Notice that if $E\subset F$, then $m_*(E)\leq m_*(F)$, and that if $E$ is meausurable, then $m_*(E)=m(E)$ by the inner regularity of $m$.

Prop 2. For any measurable set $A$ containing $E$ we have \begin{align} m(A)=m^*(E)+m_*(A\setminus E)\tag{4}\label{four} \end{align}

Proof: Suppose $G$ is man open set containing $E$. Then $$m(G)+m_*(A\setminus E) \geq m(G\cap A)+m(A\setminus G)=m(A)$$ Hence $$m^*(E)+ m_*(A\setminus E)\geq m(A)$$ Conversely, if $K$ is a compact subset of $A\setminus E$, then $E\cap A= E\subset A\setminus K$ and so, $$m^*(E)+m(K)\leq m(A\setminus K)+m(K)=m(A)$$ Hence $$m^*(E)+m_*(A\setminus E)\leq m(A)$$ Therefore $$ m^*(E)+m_*(A\setminus E)=m(A)$$

As a consequence

Prop 3. If $m^*(E)<\infty$ and $A$ is a measurable set such that $E\subset A$ and $m^*(E)=m(A)$, then $m_*(A\setminus E)=0$ and \begin{align} m^*(E\cap B)=m(A\cap B)\tag{5}\label{five} \end{align} for every measurable set $B$.

Proof: The first statement follows from \eqref{two}. Now suppose $B$ is measurable. $E\cap B\subset A\cap B$ and so \begin{align} m^*(E\cap B)&= m(A\cap B)+m_*\big((A\cap B)\setminus(E\cap B)\big)\\ &=m(A\cap B)+ m_*((A\setminus E)\cap B). \end{align} The conclusion follows from $m_*(A\setminus E)\cap B)\leq m_*(A\setminus E)=0$.

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